The rate of growth of bacteria is proportional to the bacteria present. If it is found
that the number doubles in 3 hours, then the number of times the bacteria are
increased in 6 hours is

Let \(b\) be the number of bacteria.
We have \(\frac{\mathrm{db}}{\mathrm{dt}} \propto \mathrm{b} \Rightarrow \int \frac{\mathrm{db}}{\mathrm{b}}=\int \mathrm{Kdt}\)
\(\therefore \log \mathrm{b}=\mathrm{Kt}+\mathrm{c}\) ...(1)
Let \(b_{0}\) be the initial number of bacteria. At \(t=0, b=b_{0}\)
\(\log \mathrm{b}_{0}=\mathrm{K}(0)+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{b}_{0}\)
\(\therefore \log \left(\frac{b}{b_{0}}\right)=\mathrm{Kt}\) ...(2)
When \(t=3, b=2 b_{0}\)
\(\therefore \log \left(\frac{2 b_{0}}{b_{0}}\right)=3 K \Rightarrow K=\frac{1}{3}(\log 2)\)
Thus \(\log b=\frac{1}{3}(\log 2) t+\log b_{0}\)
When \(\mathrm{t}=6\)
\(\log \left(\frac{b}{b_{0}}\right)=2 \log 2=\log 4 \Rightarrow \frac{b}{b_{0}}=4 \Rightarrow b=4 b_{0}\)
This problem can also be solved as follows :
The number of bacteria doubles in 3 hours.
Let initial number of bacteria \(=\mathrm{N}\).
\(\therefore\) After 3 hours, number of bacteria \(=2 \mathrm{~N}\).
After 6 hours, number of bacteria \(=4 \mathrm{~N}\).